For a short exact sequence $0 \to G \to H \to K \to 0$ of (discrete) groups with $K$ finite we have, as a consequence of the Hochschild-Serre spectral sequence, that $H^{\ast}(H;\mathbb Q) = H^{\ast}(G;\mathbb Q)^K$. This can be used to see that free and free abelian groups embedd with finite index into groups with trivial rational cohomology: $$H^{\ast}(F_n \rtimes_{\text{sign}} \mathbb Z/2;\mathbb Q) = H^{\ast}(\bigvee_n S^1;\mathbb Q)^{\mathbb Z/2} = H^{\ast}(\text{pt};\mathbb Q).$$ $$H^{\ast}(\mathbb Z^n \rtimes_{\text{sign}} (\mathbb Z/2)^n;\mathbb Q) = H^{\ast}(\prod_n S^1;\mathbb Q)^{(\mathbb Z/2)^n} = H^{\ast}(\text{pt};\mathbb Q).$$ Does this work for every group? Or writing down the opposite:

**Is there a group $G$ such that every group $H$ which contains $G$ with finite index has nontrivial rational cohomology?**